For each of the following set, show that they are undecidable. Do not use Rice theorem.
a. $L_{1} = \{M |M$ accepts w if w contains the substring 10 $\}$
b. $L_{2} = \{M| M$ accepts an odd number of strings $\}$
I have tried proving using aTM because it is known undecidable, but I do not know how to create the argument formally.
Suppose $L_{1}$ is decidable. Let $T$ be a TM that decides $L_{1}$. Let $M$ be a Turing Machine and $s$ be an arbitrary string. We construct $M_{s}$ which first simulates $M$ on $s$. $M_{s}$ halts if and only if $M$ accepts $s$.
We now consider $M^{\prime}$ as a Turing Machine. If $M^{\prime}$ operates by simulating $M_{s}$. $M^{\prime}$ accepts strings with $10$ as a substring if and only if $M_{s}$ halts. This implies undecidability.
Take the same approach for $L_{2}$, basically.